sm8 127

_______________________________________________________________________
PROBLEM (8.127) The 60o strain rosette (Fig. P8.127) is mounted on the surface of an
automobile frame. The following readings are obtained during a static test:
a
ε
= 1,110
μ
, b
ε
= 420
μ
, c
ε
= -240
μ
Determine:
(a) The principal strains.
(b) The maximum shear strain.
Sketch the results obtained in (a) on a properly oriented deformed element.
SOLUTION
1,110 420 240
abc
ε
με με ===− μ
a
0 60 120
oo
ab c
θθ θ ==− =−
Thus
22
cos sin sin cos
ax ay axy a
ε
εθεθγθ =++ θ
xyxy
με ε γ =++ 1,100
x
,
22
1,110 cos 0 sin 0 sin0 cos0
oooo
μ =
o
ε
Similarly,
22
420 cos ( 60 ) sin ( 60 ) sin( 60 )cos( 60 )
ooo
xyxy
με ε γ =−+−+− −
or
420 0.25 0.75 0.443
x
yxy
μ
εε =+− γ
o
(1)
Likewise,
22
240 cos ( 120 ) sin ( 120 ) sin( 120 )cos( 120 )
ooo
xyxy
με ε γ −= −+ −+ − −
or
240 0.25 0.75 0.443
x
yxy
μ
εε −= + + γ (2)
Subtract Eq. (2) from Eq. (1):
660 0.866 , 762
xy xy
μ
γγ =− =− μ
Then Eq. (2) gives 250
y
ε
μ =−
Continued on next slide
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(a)
1(
2
avg x y
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or
instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other
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)
ε
εε =+
1(1110 250) 430
2
μ
=−=
22 (1110 430) (381)R=−+
779.5
μ
=
1,2 avg
R
ε
ε =±
1430 779.5 1210
ε
μ =+ =
2430 779.5 350
ε
μ =− =−
(b) max 1 2 1210 350 1560
γ
εε μ =−= + =
Sketch:
381
tan2 ' , ' 14.6
1110 430 o
pp
θθ ==
−
1210
μ
14.6o
x’
x
y’
y
350
μ
2
ε
ε
(
μ
)
1
ε
γ
(μ)
430
2
O
C
'
2
p
θ
1110
381
R